Is there any analytical result on the following integral? $$\int_{-\infty}^{\infty} \frac{e^{-x^2}}{1+e^{-(x-\mu)}}dx$$ Thanks a lot!
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4$\begingroup$ This is a case of the Mordell integral. You might find some information in one of my questions at Mathematica.se. $\endgroup$– მამუკა ჯიბლაძეCommented Jul 15, 2016 at 4:58
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$\begingroup$ I am actually curious in what context did you encounter it. Could you share some of your motivation? $\endgroup$– მამუკა ჯიბლაძეCommented Jul 15, 2016 at 5:01
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1$\begingroup$ @მამუკაჯიბლაძე Thanks for the information. In the linear-nonlinear model of signal process of a single neuron, that integral appears very often. So far, many computation results came out. However, I wanna try to get some analytical results. $\endgroup$– hyhuCommented Jul 15, 2016 at 5:06
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2 Answers
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The integral is a special case of what is called Gauss–Fermi function in semiconductor physics. In general, it cannot be calculated analytically. However some semi-analytical approximations do exist. See, for example, http://link.springer.com/article/10.1007/s10825-014-0615-7 (Analytical evaluation of charge carrier density of organic materials with Gauss density of states, by T. Mehmetoğlu) and references therein.
answered Jul 15, 2016 at 5:25
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Maple can evaluate this integral for integer and half-integer values of $\mu$. The values seem have a nice pattern. Let the value of the integral be $F(\mu)$. Then $F(0)= \frac12\sqrt\pi$ and for integer $\mu$ with $1\le\mu\le 30$, $$ F(\mu) = - \sqrt\pi\,e^{-\mu^2} \biggl(\frac12 + \sum_{i=1}^{2\mu-1} (-1)^i e^{i^2/4}\Biggr).$$ There's also an obvious symmetry that gives $F(-\mu)+F(\mu)=\sqrt\pi$.
For half-integers, the empirical pattern is $F(\mu+\frac12)=F(-\mu)e^{-\mu-1/4}$ (checked for lots of integer $\mu$).
I assume these are all easy to prove but I'll leave that for someone else.
answered Jul 15, 2016 at 7:25